CBSE Class 9 Mathematics Herons Formula Assignment Set 05

Question 1. Write Heron’s formula to find the area of a triangle.
Answer: The area of a triangle with sides \( a \), \( b \), and \( c \) is determined using the formula \( \text{Area} = \sqrt{s(s-a)(s-b)(s-c)} \), where the variable \( s \) represents the semi-perimeter: \[ s = \frac{a+b+c}{2} \]
In simple words: This formula helps you find a triangle's area using just its three side lengths. First, add the sides and divide by two, then plug that value into the square root equation.
Exam Tip: Always state the definition of \( s \) clearly in your steps to secure full marks for formula presentation.

Question 2. Write the area of the rhombus , if \( d_1 \) and \( d_2 \) are the lengths of its diagonals .
Answer: When the diagonal lengths of a rhombus are given as \( d_1 \) and \( d_2 \), its total area is calculated using the relation: \[ \text{Area} = \frac{1}{2} \times d_1 \times d_2 \]
In simple words: To find the area of a rhombus, multiply the lengths of its two diagonals together and then divide the result by two.
Exam Tip: Ensure that both diagonal lengths are in the same units before multiplying them.

Question 3. What is the area of equilateral triangle whose side is a units ?
Answer: For an equilateral triangle with a side length of \( a \) units, the formula to find the area is: \[ \text{Area} = \frac{\sqrt{3}}{4} a^2 \text{ square units} \]
In simple words: When all three sides of a triangle are equal to \( a \), you can find its area directly by squaring \( a \) and multiplying it by \( \frac{\sqrt{3}}{4} \).
Exam Tip: Do not forget to write "square units" or the appropriate metric units squared in your final answer.

Question 4. What is the area of an isosceles right angled triangle whose equal side is a units ?
Answer: Since the two equal sides of an isosceles right-angled triangle are perpendicular to each other, they serve as the base and the height. Therefore, the area is: \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} a^2 \text{ square units} \]
In simple words: If a right triangle has two equal sides of length \( a \), its area is just half of \( a \) squared.
Exam Tip: Remember that the hypotenuse is the longest side, so the two equal sides are always the ones forming the right angle.

Question 5. What is the side of a rhombus whose diagonal is \( d_1 \) and \( d_2 \)?
Answer: The diagonals of a rhombus bisect each other at right angles. Applying the Pythagorean theorem to one of the four internal triangles, the side length is: \[ \text{Side} = \sqrt{\left(\frac{d_1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2} = \frac{1}{2} \sqrt{d_1^2 + d_2^2} \]
In simple words: You can find the outer side of a rhombus by taking half of the square root of the sum of both squared diagonals.
Exam Tip: Deriving this relation using a neat diagram of a rhombus helps show the examiner your conceptual clarity.

Question 6. Find the area of a triangle whose sides are 13cm, 14cm and 15cm.
Answer: Let the sides of the triangle be \( a = 13 \text{ cm} \), \( b = 14 \text{ cm} \), and \( c = 15 \text{ cm} \).
First, find the semi-perimeter \( s \): \[ s = \frac{13 + 14 + 15}{2} = \frac{42}{2} = 21 \text{ cm} \]
Now, use Heron’s formula to determine the area: \[ \text{Area} = \sqrt{s(s-a)(s-b)(s-c)} \]
\( \implies \text{Area} = \sqrt{21(21-13)(21-14)(21-15)} \)
\( \implies \text{Area} = \sqrt{21 \times 8 \times 7 \times 6} \)
\( \implies \text{Area} = \sqrt{3 \times 7 \times 2^3 \times 7 \times 2 \times 3} \)
\( \implies \text{Area} = \sqrt{7^2 \times 3^2 \times 2^4} \)
\( \implies \text{Area} = 7 \times 3 \times 4 = 84 \text{ cm}^2 \)
In simple words: Calculate the semi-perimeter (21) first, subtract each side length from it, multiply those differences with 21, and find the square root to get 84.
Exam Tip: Grouping terms into prime factors under the square root makes calculation faster and prevents multiplication mistakes.

Question 7. The perimeter of a triangular field is 450 m and its sides are in the ratio 13:12:15. Find the area of the triangle.
Answer: Let the three sides of the triangular field be \( 13x \), \( 12x \), and \( 15x \) meters.
Using the given perimeter: \[ 13x + 12x + 15x = 450 \]
\( \implies 40x = 450 \)
\( \implies x = 11.25 \text{ m} \)
Thus, the side lengths are:
\( a = 13 \times 11.25 = 146.25 \text{ m} \)
\( b = 12 \times 11.25 = 135.00 \text{ m} \)
\( c = 15 \times 11.25 = 168.75 \text{ m} \)
The semi-perimeter \( s \) is: \[ s = \frac{450}{2} = 225 \text{ m} \]
Applying Heron's formula: \[ \text{Area} = \sqrt{s(s-a)(s-b)(s-c)} \]
\( \implies \text{Area} = \sqrt{225(225-146.25)(225-135)(225-168.75)} \)
\( \implies \text{Area} = \sqrt{225 \times 78.75 \times 90 \times 56.25} \)
\( \implies \text{Area} = \sqrt{15^2 \times \frac{315}{4} \times 90 \times \frac{225}{4}} \)
\( \implies \text{Area} = \sqrt{\frac{225^2 \times 28350}{16}} \)
\( \implies \text{Area} = \frac{225 \times 45\sqrt{14}}{4} = \frac{10125\sqrt{14}}{4} \approx 9471.07 \text{ m}^2 \)
In simple words: First find the actual side lengths from the ratio, then find the semi-perimeter, and finally use the square-root formula to calculate the area.
Exam Tip: When decimal values occur, keep fractional forms like \( \frac{315}{4} \) under the root to make simplification easier.

Question 8. Find the area of a triangle whose two sides are 8 cm and 11cm and the perimeter is 32cm.
Answer: Let the sides be \( a = 8 \text{ cm} \) and \( b = 11 \text{ cm} \).
The perimeter is \( 32 \text{ cm} \), so the third side \( c \) is: \[ c = 32 - (8 + 11) = 32 - 19 = 13 \text{ cm} \]
The semi-perimeter \( s \) is: \[ s = \frac{32}{2} = 16 \text{ cm} \]
Now, apply Heron's formula: \[ \text{Area} = \sqrt{s(s-a)(s-b)(s-c)} \]
\( \implies \text{Area} = \sqrt{16(16-8)(16-11)(16-13)} \)
\( \implies \text{Area} = \sqrt{16 \times 8 \times 5 \times 3} \)
\( \implies \text{Area} = \sqrt{16 \times 120} = 8\sqrt{30} \text{ cm}^2 \approx 43.82 \text{ cm}^2 \)
In simple words: Subtract the two known sides from the total perimeter to get the third side (13 cm), then find the area using Heron's formula.
Exam Tip: Leave the final answer in square root form (\( 8\sqrt{30} \)) unless the value of \( \sqrt{30} \) is explicitly given in the paper.

Question 9. The lengths of the sides of a triangle are 5 cm, 12 cm and 13 cm. Find the length of the perpendicular from the opposite vertex to the side whose length is 13 cm.
Answer: First, observe that the side lengths \( 5 \text{ cm} \), \( 12 \text{ cm} \), and \( 13 \text{ cm} \) satisfy the Pythagorean theorem: \[ 5^2 + 12^2 = 25 + 144 = 169 = 13^2 \]
This means the triangle is right-angled, with base \( = 12 \text{ cm} \) and height \( = 5 \text{ cm} \).
The area of the triangle is: \[ \text{Area} = \frac{1}{2} \times 12 \times 5 = 30 \text{ cm}^2 \]
Let the perpendicular to the longest side (13 cm) be \( h \). Using the same area: \[ \text{Area} = \frac{1}{2} \times 13 \times h \]
\( \implies 30 = \frac{13}{2} h \)
\( \implies h = \frac{60}{13} \approx 4.62 \text{ cm} \)
In simple words: First calculate the triangle's area (30) since it is a right triangle. Then, use the 13 cm side as the base and solve for the height to get \( 4.62 \text{ cm} \).
Exam Tip: Recognizing Pythagorean triplets right away saves you from using the longer Heron's formula to find the area.

Question 10. If the sides of the triangle are 26 cm, 28 cm and 30 cm, find the area of the triangle.
Answer: Let the sides be \( a = 26 \text{ cm} \), \( b = 28 \text{ cm} \), and \( c = 30 \text{ cm} \).
First, find the semi-perimeter \( s \): \[ s = \frac{26 + 28 + 30}{2} = \frac{84}{2} = 42 \text{ cm} \]
Applying Heron's formula: \[ \text{Area} = \sqrt{s(s-a)(s-b)(s-c)} \]
\( \implies \text{Area} = \sqrt{42(42-26)(42-28)(42-30)} \)
\( \implies \text{Area} = \sqrt{42 \times 16 \times 14 \times 12} \)
\( \implies \text{Area} = \sqrt{(14 \times 3) \times 16 \times 14 \times (3 \times 4)} \)
\( \implies \text{Area} = \sqrt{14^2 \times 3^2 \times 16 \times 4} \)
\( \implies \text{Area} = 14 \times 3 \times 4 \times 2 = 336 \text{ cm}^2 \)
In simple words: Find the semi-perimeter (42), subtract each side from it, multiply everything together under the root, and simplify to get 336.
Exam Tip: Breaking down large numbers like 42 and 12 into smaller factors under the radical sign makes finding the square root much easier.

Question 11. The perimeter of a right triangle is 450 m. If its sides are in the ratio 13 : 12 : 5. Find the area of the triangle.
Answer: Let the side lengths of the right triangle be \( 13x \), \( 12x \), and \( 5x \) meters.
Given that the perimeter is 450 m: \[ 13x + 12x + 5x = 450 \]
\( \implies 30x = 450 \)
\( \implies x = 15 \]
The sides are:
Hypotenuse \( = 13 \times 15 = 195 \text{ m} \)
Base/Height sides \( = 12 \times 15 = 180 \text{ m} \) and \( 5 \times 15 = 75 \text{ m} \)
Since it is a right-angled triangle, the area is: \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]
\( \implies \text{Area} = \frac{1}{2} \times 180 \times 75 = 90 \times 75 = 6750 \text{ m}^2 \)
In simple words: Find the value of \( x \) (15) to get the actual sides, and then calculate the area using the formula for a right triangle.
Exam Tip: For right-angled triangles, use the simple \( \frac{1}{2} \times \text{base} \times \text{height} \) formula instead of Heron's formula to save valuable exam time.

Question 12. There is a slide in a park. One of its side walls has been painted in some colours with a message “KEEP THE PARK GREEN AND CLEAN”. If the sides of the wall are 15m, 11m and 6m, find the area painted in colour.
Answer: The painted wall forms a triangle with sides \( a = 15 \text{ m} \), \( b = 11 \text{ m} \), and \( c = 6 \text{ m} \).
The semi-perimeter \( s \) is: \[ s = \frac{15 + 11 + 6}{2} = \frac{32}{2} = 16 \text{ m} \]
Using Heron's formula: \[ \text{Area} = \sqrt{s(s-a)(s-b)(s-c)} \]
\( \implies \text{Area} = \sqrt{16(16-15)(16-11)(16-6)} \)
\( \implies \text{Area} = \sqrt{16 \times 1 \times 5 \times 10} \)
\( \implies \text{Area} = \sqrt{800} = \sqrt{400 \times 2} = 20\sqrt{2} \text{ m}^2 \approx 28.28 \text{ m}^2 \)
In simple words: Find the semi-perimeter of the triangular wall, and then use the area formula to get the painted space, which is \( 20\sqrt{2} \text{ m}^2 \).
Exam Tip: Be sure to write a concluding statement that relates your mathematical answer back to the real-world scenario described in the question.

Question 13. If the side of an equilateral triangle is ‘a’, then find the altitude of the equilateral triangle.
Answer: Let the equilateral triangle be \( \Delta ABC \) with side length \( a \). The altitude drawn from vertex \( A \) to base \( BC \) bisects \( BC \) at point \( D \), making \( BD = \frac{a}{2} \).
In the right-angled triangle \( \Delta ABD \), using the Pythagorean theorem: \[ AB^2 = AD^2 + BD^2 \]
\( \implies a^2 = AD^2 + \left(\frac{a}{2}\right)^2 \)
\( \implies AD^2 = a^2 - \frac{a^2}{4} = \frac{3a^2}{4} \)
\( \implies AD = \frac{\sqrt{3}}{2} a \)
In simple words: Splitting an equilateral triangle down the middle gives you a right-angled triangle. Solving for the height gives \( \frac{\sqrt{3}}{2} a \).
Exam Tip: Draw a clean diagram showing the altitude bisecting the base to help support your algebraic proof.

Question 14. Find the area of a triangle whose two sides are 8 cm and 11cm and the perimeter is 32cm.
Answer: Let the sides of the triangle be \( a = 8 \text{ cm} \) and \( b = 11 \text{ cm} \).
The perimeter is \( 32 \text{ cm} \), so the third side \( c \) is: \[ c = 32 - (8 + 11) = 32 - 19 = 13 \text{ cm} \]
The semi-perimeter \( s \) is: \[ s = \frac{32}{2} = 16 \text{ cm} \]
Applying Heron's formula: \[ \text{Area} = \sqrt{s(s-a)(s-b)(s-c)} \]
\( \implies \text{Area} = \sqrt{16(16-8)(16-11)(16-13)} \)
\( \implies \text{Area} = \sqrt{16 \times 8 \times 5 \times 3} \)
\( \implies \text{Area} = \sqrt{16 \times 4 \times 2 \times 5 \times 3} = 4 \times 2 \sqrt{30} = 8\sqrt{30} \text{ cm}^2 \approx 43.82 \text{ cm}^2 \)
In simple words: Find the third side first by subtracting 8 and 11 from the perimeter, and then use Heron's formula to get \( 8\sqrt{30} \text{ cm}^2 \).
Exam Tip: Carefully check all addition and subtraction steps when finding the third side of a triangle.

Question 15. A rhombus shaped field has green grass for 18 horses to graze. If each side of the rhombus is 30m and its longer diagonal is 48m, how much area of grass field will each Horse be grazing?
Answer: The diagonal of a rhombus divides it into two congruent triangles. Each triangle has side lengths \( a = 30 \text{ m} \), \( b = 30 \text{ m} \), and \( c = 48 \text{ m} \).
For one of these triangles, the semi-perimeter \( s \) is: \[ s = \frac{30 + 30 + 48}{2} = \frac{108}{2} = 54 \text{ m} \]
The area of one triangle is: \[ \text{Area} = \sqrt{54(54-30)(54-30)(54-48)} \]
\( \implies \text{Area} = \sqrt{54 \times 24 \times 24 \times 6} = \sqrt{9 \times 6 \times 24 \times 24 \times 6} = 3 \times 6 \times 24 = 432 \text{ m}^2 \)
The total area of the rhombus is: \[ \text{Total Area} = 2 \times 432 = 864 \text{ m}^2 \]
Since there are 18 horses, the area available for each horse is: \[ \text{Area per Horse} = \frac{864}{18} = 48 \text{ m}^2 \)
In simple words: The diagonal splits the field into two identical triangles. Find the total area (864) and divide it by the 18 horses to get 48 square meters each.
Exam Tip: Be sure to divide the final total area by the number of horses to answer the exact question asked.

Question 16. Find the area of a quadrilateral ABCD whose sides are 9m, 40m, 28m and 15m respectively and the angle between the first two sides is a right angle.
Answer: Let the quadrilateral be ABCD where \( AB = 9 \text{ m} \), \( BC = 40 \text{ m} \), \( CD = 28 \text{ m} \), and \( DA = 15 \text{ m} \). The angle between AB and BC is \( 90^\circ \).
In right-angled triangle \( \Delta ABC \), the hypotenuse \( AC \) is: \[ AC = \sqrt{AB^2 + BC^2} = \sqrt{9^2 + 40^2} = \sqrt{81 + 1600} = \sqrt{1681} = 41 \text{ m} \]
The area of \( \Delta ABC \) is: \[ \text{Area}(\Delta ABC) = \frac{1}{2} \times 9 \times 40 = 180 \text{ m}^2 \]
For the second triangle \( \Delta ADC \), the side lengths are \( 15 \text{ m} \), \( 28 \text{ m} \), and \( 41 \text{ m} \).
Its semi-perimeter \( s \) is: \[ s = \frac{15 + 28 + 41}{2} = \frac{84}{2} = 42 \text{ m} \]
The area of \( \Delta ADC \) is: \[ \text{Area}(\Delta ADC) = \sqrt{42(42-15)(42-28)(42-41)} \]
\( \implies \text{Area}(\Delta ADC) = \sqrt{42 \times 27 \times 14 \times 1} = \sqrt{14 \times 3 \times 27 \times 14} = 14 \times 9 = 126 \text{ m}^2 \)
The total area of quadrilateral ABCD is: \[ \text{Total Area} = \text{Area}(\Delta ABC) + \text{Area}(\Delta ADC) = 180 + 126 = 306 \text{ m}^2 \)
In simple words: Split the shape into two triangles using a diagonal line. Find the area of both parts separately, then add them together to get 306.
Exam Tip: Splitting quadrilaterals into two triangles is the standard way to solve area problems using Heron's formula.

Question 17. Find the area of a rhombus whose perimeter is 80m and one of diagonal is 24m.
Answer: Let the side of the rhombus be \( a \).
The perimeter is 80 m, so: \[ a = \frac{80}{4} = 20 \text{ m} \]
Let the diagonals be \( d_1 = 24 \text{ m} \) and \( d_2 \).
The diagonals of a rhombus bisect each other perpendicularly, so: \[ \left(\frac{d_1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2 = a^2 \]
\( \implies 12^2 + \left(\frac{d_2}{2}\right)^2 = 20^2 \)
\( \implies 144 + \left(\frac{d_2}{2}\right)^2 = 400 \)
\( \implies \left(\frac{d_2}{2}\right)^2 = 256 \implies \frac{d_2}{2} = 16 \implies d_2 = 32 \text{ m} \)
The area of the rhombus is: \[ \text{Area} = \frac{1}{2} \times d_1 \times d_2 = \frac{1}{2} \times 24 \times 32 = 384 \text{ m}^2 \)
In simple words: Find the side length (20) first. Use the right-triangle rule inside the rhombus to get the second diagonal (32), then calculate the final area (384).
Exam Tip: You can also solve this by finding the area of one triangle with sides 20 m, 20 m, and 24 m using Heron's formula, then doubling it.

Question 18. A floral design on a floor is made up of 16 tiles that are triangular, the sides of the triangle being 9cm, 28cm and 35cm. Find the area of the floral design.
Answer: For each of the 16 identical triangular tiles, let the sides be \( a = 9 \text{ cm} \), \( b = 28 \text{ cm} \), and \( c = 35 \text{ cm} \).
The semi-perimeter \( s \) is: \[ s = \frac{9 + 28 + 35}{2} = \frac{72}{2} = 36 \text{ cm} \]
The area of a single tile is: \[ \text{Area of 1 tile} = \sqrt{36(36-9)(36-28)(36-35)} \]
\( \implies \text{Area of 1 tile} = \sqrt{36 \times 27 \times 8 \times 1} \)
\( \implies \text{Area of 1 tile} = \sqrt{36 \times 216} = 6 \sqrt{216} = 36\sqrt{6} \text{ cm}^2 \approx 36 \times 2.45 = 88.2 \text{ cm}^2 \)
The total area of all 16 tiles is: \[ \text{Total Area} = 16 \times 36\sqrt{6} = 576\sqrt{6} \text{ cm}^2 \approx 1410.9 \text{ cm}^2 \br />In simple words: Calculate the area of just one triangle first (\( 36\sqrt{6} \)), and then multiply that by 16 to get the total area.
Exam Tip: Be careful with the multiplication steps when scaling up the area from one tile to multiple tiles.

Question 19. Compute the area of trapezium.
Answer: Consider the trapezium ABCD as shown in the diagram.
Let \( CO \) be the perpendicular drawn from \( C \) to the base \( AB \).
In the right-angled triangle \( \Delta COB \), the hypotenuse is \( BC = 17 \text{ cm} \) and the base is \( OB = 8 \text{ cm} \).
Using the Pythagorean theorem: \[ CO^2 + OB^2 = BC^2 \]
\( \implies CO^2 + 8^2 = 17^2 \)
\( \implies CO^2 + 64 = 289 \)
\( \implies CO^2 = 225 \implies CO = 15 \text{ cm} \)
The height \( h \) of the trapezium is \( 15 \text{ cm} \).
The parallel sides of the trapezium are \( CD = 6 \text{ cm} \) and \( AB = AO + OB = 6 + 8 = 14 \text{ cm} \).
The area of the trapezium is: \[ \text{Area} = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height} \]
\( \implies \text{Area} = \frac{1}{2} \times (6 + 14) \times 15 = \frac{1}{2} \times 20 \times 15 = 150 \text{ cm}^2 \) In simple words: First find the height of the shape (15) using the right-angled triangle on the right side. Then use the formula for a trapezium to get 150.
Exam Tip: Be sure to add the segment lengths AO and OB to get the entire bottom base length AB.

Question 20. In fig. given below, BD is the diagonal of quadrilateral ABCD. Find the area of ABCD.
Answer: The quadrilateral ABCD is split by its diagonal BD into two right-angled triangles: \( \Delta ABD \) and \( \Delta BDC \).
For \( \Delta ABD \), the base is \( AB = 5 \text{ cm} \) and the height is \( BD = 7 \text{ cm} \). \[ \text{Area}(\Delta ABD) = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 \times 7 = 17.5 \text{ cm}^2 \]
For \( \Delta BDC \), the base is \( CD = 5 \text{ cm} \) and the height is \( BD = 7 \text{ cm} \). \[ \text{Area}(\Delta BDC) = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 \times 7 = 17.5 \text{ cm}^2 \]
The total area of quadrilateral ABCD is: \[ \text{Total Area} = \text{Area}(\Delta ABD) + \text{Area}(\Delta BDC) = 17.5 + 17.5 = 35 \text{ cm}^2 \] In simple words: The diagonal splits the shape into two equal right-angled triangles. Each has an area of 17.5, so adding them together gives 35.
Exam Tip: Look for right angle symbols in the drawing to identify where you can apply simple right-triangle formulas.

Question 21. Compute the area of the following trapezium:
Answer: Let \( CO \) be the vertical height of the trapezium.
In the right triangle \( \Delta COB \), using the Pythagorean theorem with hypotenuse \( BC = 17 \text{ cm} \) and base segment \( OB = 8 \text{ cm} \): \[ CO^2 + OB^2 = BC^2 \]
\( \implies CO^2 + 8^2 = 17^2 \)
\( \implies CO^2 + 64 = 289 \implies CO = 15 \text{ cm} \)
The parallel sides are \( CD = 6 \text{ cm} \) (since \( CD = AO \)) and the full base \( AB = AO + OB = 6 + 8 = 14 \text{ cm} \).
The area is: \[ \text{Area} = \frac{1}{2} \times (CD + AB) \times CO \]
\( \implies \text{Area} = \frac{1}{2} \times (6 + 14) \times 15 = \frac{1}{2} \times 20 \times 15 = 150 \text{ cm}^2 \) In simple words: This uses the same steps as Question 19. Find the height (15) first, then calculate the total area which comes to 150.
Exam Tip: Be sure to write down the steps of the Pythagorean theorem clearly as it carries separate method marks.

Question 22. Find area and perimeter of triangle whose sides are 8cm ,19cm and 15 cm.
Answer: Let the side lengths be \( a = 8 \text{ cm} \), \( b = 19 \text{ cm} \), and \( c = 15 \text{ cm} \).
The perimeter is: \[ \text{Perimeter} = a + b + c = 8 + 19 + 15 = 42 \text{ cm} \]
The semi-perimeter \( s \) is: \[ s = \frac{42}{2} = 21 \text{ cm} \]
Applying Heron's formula: \[ \text{Area} = \sqrt{s(s-a)(s-b)(s-c)} \]
\( \implies \text{Area} = \sqrt{21(21-8)(21-19)(21-15)} \)
\( \implies \text{Area} = \sqrt{21 \times 13 \times 2 \times 6} \)
\( \implies \text{Area} = \sqrt{3276} = \sqrt{36 \times 91} = 6\sqrt{91} \text{ cm}^2 \approx 57.24 \text{ cm}^2 \)
In simple words: Add the sides to find the perimeter (42). Then, use the semi-perimeter (21) in Heron's formula to find the area, which is \( 6\sqrt{91} \text{ cm}^2 \).
Exam Tip: If the root doesn't simplify to an integer, leaving it as a simplified surd (\( 6\sqrt{91} \)) is perfectly correct.

Question 23. Find the area of triangle whose sides are 5cm,12cm, 13 cm . Also find the Shortest altitude.
Answer: The sides \( 5 \text{ cm} \), \( 12 \text{ cm} \), and \( 13 \text{ cm} \) form a right-angled triangle because \( 5^2 + 12^2 = 13^2 \).
The area of the triangle is: \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 12 \times 5 = 30 \text{ cm}^2 \]
The shortest altitude is the perpendicular drawn to the longest side (the hypotenuse, 13 cm). Let this altitude be \( h \): \[ \text{Area} = \frac{1}{2} \times 13 \times h \]
\( \implies 30 = \frac{13}{2} h \implies h = \frac{60}{13} \approx 4.62 \text{ cm} \)
In simple words: Calculate the area (30) of the right-angled triangle first. To find the shortest height, use the hypotenuse (13) as the base.
Exam Tip: Remember that the shortest altitude of any triangle is always dropped onto its longest side.

Question 24. In a rectangular field of dimension 50 ft x 30 ft, a triangular park is constructed. If the dimension of the triangular park is 14 ft, 15 ft and 13 ft, find the area of the remaining field.
Answer: First, calculate the total area of the rectangular field: \[ \text{Area of Rectangle} = 50 \times 30 = 1500 \text{ sq ft} \]
Now, find the area of the triangular park with sides \( a = 14 \text{ ft} \), \( b = 15 \text{ ft} \), and \( c = 13 \text{ ft} \).
Its semi-perimeter \( s \) is: \[ s = \frac{14 + 15 + 13}{2} = \frac{42}{2} = 21 \text{ ft} \]
Using Heron's formula: \[ \text{Area of Triangle} = \sqrt{s(s-a)(s-b)(s-c)} \]
\( \implies \text{Area of Triangle} = \sqrt{21(21-14)(21-15)(21-13)} \)
\( \implies \text{Area of Triangle} = \sqrt{21 \times 7 \times 6 \times 8} \)
\( \implies \text{Area of Triangle} = \sqrt{7056} = 84 \text{ sq ft} \)
Subtract the triangular area from the rectangular area to find the remaining space: \[ \text{Remaining Area} = 1500 - 84 = 1416 \text{ sq ft} \]
In simple words: Find the area of the rectangle (1500) and the triangle (84) separately. Subtract the triangle's area from the rectangle's area to get 1416.
Exam Tip: Always subtract the smaller area from the larger one, and double-check your subtraction steps.

Question 25. Sanya owns a piece of land which is in the shape of a rhombus. She wants her daughter and son to work on the land and produce different crops to suffice the needs of their family. She divided the land in two equal parts. If the perimeter of the land is 400 m and one of the diagonals is 160 m, how much area each of them will get?
Answer: Since the land is a rhombus with a perimeter of 400 m, each side \( a \) is: \[ a = \frac{400}{4} = 100 \text{ m} \]
The diagonal of 160 m divides the rhombus into two identical triangles. Each child gets one triangle with sides \( 100 \text{ m} \), \( 100 \text{ m} \), and \( 160 \text{ m} \).
For one triangle, the semi-perimeter \( s \) is: \[ s = \frac{100 + 100 + 160}{2} = \frac{360}{2} = 180 \text{ m} \]
The area each child gets is: \[ \text{Area} = \sqrt{s(s-a)(s-b)(s-c)} \]
\( \implies \text{Area} = \sqrt{180(180-100)(180-100)(180-160)} \)
\( \implies \text{Area} = \sqrt{180 \times 80 \times 80 \times 20} \)
\( \implies \text{Area} = 80 \times \sqrt{180 \times 20} = 80 \times \sqrt{3600} = 80 \times 60 = 4800 \text{ m}^2 \)
In simple words: The land is split into two equal triangles with sides 100, 100, and 160. Calculating the area of one triangle gives 4800 square meters for each child.
Exam Tip: Make sure to state that the diagonal of a rhombus divides it into two congruent triangles of equal area.

Question 26. An umbrella is made by stitching 10 triangular pieces of cloth of two different colours, each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for the umbrella?
Answer: The umbrella consists of 10 triangular pieces of cloth, which means there are 5 pieces of each of the two colors.
For one triangular piece, let the sides be \( a = 20 \text{ cm} \), \( b = 50 \text{ cm} \), and \( c = 50 \text{ cm} \).
The semi-perimeter \( s \) is: \[ s = \frac{20 + 50 + 50}{2} = \frac{120}{2} = 60 \text{ cm} \]
The area of one piece is: \[ \text{Area} = \sqrt{s(s-a)(s-b)(s-c)} \]
\( \implies \text{Area} = \sqrt{60(60-20)(60-50)(60-50)} \)
\( \implies \text{Area} = \sqrt{60 \times 40 \times 10 \times 10} \)
\( \implies \text{Area} = 10 \times \sqrt{2400} = 10 \times 20\sqrt{6} = 200\sqrt{6} \text{ cm}^2 \)
Since there are 5 pieces of each color, the total area for each color is: \[ \text{Total Area for each color} = 5 \times 200\sqrt{6} = 1000\sqrt{6} \text{ cm}^2 \approx 2449.49 \text{ cm}^2 \br />In simple words: Find the area of one triangular cloth piece (\( 200\sqrt{6} \)). Since there are two colors, multiply this by 5 to find the total cloth needed for each color.
Exam Tip: Always read carefully to see if the question asks for the total cloth or the cloth of each color separately.

Question 27. Find the area of triangle whose sides are 5cm, 12cm, 13 cm. Also find the shortest altitude.
Answer: Let the side lengths be \( a = 5 \text{ cm} \), \( b = 12 \text{ cm} \), and \( c = 13 \text{ cm} \).
Because \( 5^2 + 12^2 = 25 + 144 = 169 = 13^2 \), these side lengths form a right-angled triangle.
The area of the triangle is: \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 12 \times 5 = 30 \text{ cm}^2 \]
The shortest altitude is perpendicular to the hypotenuse (13 cm). Let this height be \( h \): \[ \text{Area} = \frac{1}{2} \times 13 \times h \]
\( \implies 30 = \frac{13}{2} h \implies h = \frac{60}{13} \approx 4.62 \text{ cm} \)
In simple words: This triangle has an area of 30. Dividing double the area by the longest side (13) gives the shortest altitude, which is about 4.62 cm.
Exam Tip: Drawing the perpendicular height to the hypotenuse can help you visualize the geometry of this question.

Question 28. A kite is in the shape of a square with diagonal 32 cm and an isosceles triangle of base 8cm and equal sides are 6cm. How much paper is required to build the kite?
Answer: The kite is composed of two main parts: a square and a bottom triangle.
1. For the square with diagonal \( d = 32 \text{ cm} \): \[ \text{Area of Square} = \frac{1}{2} d^2 = \frac{1}{2} \times 32^2 = \frac{1}{2} \times 1024 = 512 \text{ cm}^2 \]
2. For the bottom isosceles triangle with base \( b = 8 \text{ cm} \) and sides \( a = 6 \text{ cm} \), \( c = 6 \text{ cm} \):
Its semi-perimeter \( s \) is: \[ s = \frac{6 + 6 + 8}{2} = 10 \text{ cm} \]
The area of this triangle is: \[ \text{Area} = \sqrt{s(s-a)(s-b)(s-c)} \]
\( \implies \text{Area} = \sqrt{10(10-6)(10-6)(10-8)} \)
\( \implies \text{Area} = \sqrt{10 \times 4 \times 4 \times 2} = 4\sqrt{20} = 8\sqrt{5} \text{ cm}^2 \approx 17.89 \text{ cm}^2 \)
The total paper required is the sum of both areas: \[ \text{Total Paper} = 512 + 8\sqrt{5} \text{ cm}^2 \approx 529.89 \text{ cm}^2 \] In simple words: Find the area of the square body (512) and the tail triangle (17.89) separately, then add them to find the total paper needed.
Exam Tip: Be sure to write out the calculations for both sections clearly before adding them together.

Question 29. Find the area of the trapezium whose parallel sides are 25 cm, 13 cm and other sides are 15 cm and 15 cm.
Answer: Let the parallel sides of the trapezium be \( a = 13 \text{ cm} \) and \( b = 25 \text{ cm} \), and the non-parallel sides be \( c = d = 15 \text{ cm} \).
If we draw perpendiculars from the ends of the top parallel side to the base, they divide the base into three parts. The middle part is \( 13 \text{ cm} \), and the two end segments are equal in length: \[ \text{Length of each end segment} = \frac{25 - 13}{2} = 6 \text{ cm} \]
In the right-angled triangle formed at either side, the base is \( 6 \text{ cm} \) and the hypotenuse is \( 15 \text{ cm} \). The height \( h \) of the trapezium is: \[ h = \sqrt{15^2 - 6^2} = \sqrt{225 - 36} = \sqrt{189} = 3\sqrt{21} \text{ cm} \approx 13.75 \text{ cm} \]
The area of the trapezium is: \[ \text{Area} = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height} \]
\( \implies \text{Area} = \frac{1}{2} \times (13 + 25) \times 3\sqrt{21} = 19 \times 3\sqrt{21} = 57\sqrt{21} \text{ cm}^2 \approx 261.2 \text{ cm}^2 \)
In simple words: Draw height lines to form two right-angled triangles. Use the Pythagorean theorem to find the height (\( 3\sqrt{21} \)), then find the trapezium's area.
Exam Tip: For symmetric (isosceles) trapeziums, drawing two perpendiculars is the standard approach to find the missing height.

Question 30. The side of a quadrilateral taken in order are 5,12,14 and 15 metres respectively and the angle formed by the first two sides is a right angles find its area.
Answer: Let ABCD be the quadrilateral with \( AB = 5 \text{ m} \), \( BC = 12 \text{ m} \), \( CD = 14 \text{ m} \), and \( DA = 15 \text{ m} \). The angle between AB and BC is \( 90^\circ \).
In the right-angled triangle \( \Delta ABC \), the diagonal \( AC \) is: \[ AC = \sqrt{AB^2 + BC^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \text{ m} \]
The area of \( \Delta ABC \) is: \[ \text{Area}(\Delta ABC) = \frac{1}{2} \times 5 \times 12 = 30 \text{ m}^2 \]
For the second triangle \( \Delta ADC \), the sides are \( 13 \text{ m} \), \( 14 \text{ m} \), and \( 15 \text{ m} \).
Its semi-perimeter \( s \) is: \[ s = \frac{13 + 14 + 15}{2} = 21 \text{ m} \]
Using Heron's formula: \[ \text{Area}(\Delta ADC) = \sqrt{21(21-13)(21-14)(21-15)} \]
\( \implies \text{Area}(\Delta ADC) = \sqrt{21 \times 8 \times 7 \times 6} = \sqrt{7056} = 84 \text{ m}^2 \)
The total area of the quadrilateral ABCD is: \[ \text{Total Area} = \text{Area}(\Delta ABC) + \text{Area}(\Delta ADC) = 30 + 84 = 114 \text{ m}^2 \] In simple words: The diagonal (13) splits the quadrilateral into two triangles. Find their separate areas (30 and 84) and add them together to get 114.
Exam Tip: Be sure to write down the units in meters squared (\( \text{m}^2 \)) since the dimensions are given in meters.

Question 31. Find the area and perimeter of triangle whose sides are 8cm, 19cm and 15 cm.
Answer: Let the side lengths be \( a = 8 \text{ cm} \), \( b = 19 \text{ cm} \), and \( c = 15 \text{ cm} \).
The total boundary length (perimeter) is: \[ \text{Perimeter} = 8 + 19 + 15 = 42 \text{ cm} \]
The semi-perimeter \( s \) is: \[ s = \frac{42}{2} = 21 \text{ cm} \]
Applying Heron's formula: \[ \text{Area} = \sqrt{s(s-a)(s-b)(s-c)} \]
\( \implies \text{Area} = \sqrt{21(21-8)(21-19)(21-15)} \)
\( \implies \text{Area} = \sqrt{21 \times 13 \times 2 \times 6} \)
\( \implies \text{Area} = \sqrt{3276} = 6\sqrt{91} \text{ cm}^2 \approx 57.24 \text{ cm}^2 \br />In simple words: Add up the three side lengths to get a perimeter of 42. Using Heron's formula with \( s = 21 \), the area is \( 6\sqrt{91} \text{ cm}^2 \).
Exam Tip: Double-check the sum of the sides to ensure your semi-perimeter value is correct before continuing with Heron's formula.

Question 32. Find the area of a quadrilateral ABCD which AD = 24 cm, BAD = 90° and BCD is an equilateral triangle whose each side is 26cm.
Answer: In quadrilateral ABCD, \( \Delta BCD \) is an equilateral triangle with side lengths of 26 cm, so \( BD = 26 \text{ cm} \).
In the right-angled triangle \( \Delta ABD \), we are given \( AD = 24 \text{ cm} \) and hypotenuse \( BD = 26 \text{ cm} \).
Using the Pythagorean theorem: \[ AB^2 + AD^2 = BD^2 \]
\( \implies AB^2 + 24^2 = 26^2 \)
\( \implies AB^2 + 576 = 676 \implies AB^2 = 100 \implies AB = 10 \text{ cm} \)
Now, calculate the area of both triangles:
1. For right-angled \( \Delta ABD \): \[ \text{Area}(\Delta ABD) = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10 \times 24 = 120 \text{ cm}^2 \]
2. For equilateral \( \Delta BCD \) with side \( a = 26 \text{ cm} \): \[ \text{Area}(\Delta BCD) = \frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} \times 26^2 = 169\sqrt{3} \text{ cm}^2 \approx 292.7 \text{ cm}^2 \]
The total area is the sum of both triangular areas: \[ \text{Total Area} = 120 + 169\sqrt{3} \text{ cm}^2 \approx 412.7 \text{ cm}^2 \br />In simple words: Use the right-triangle rule to find side AB (10). Then find the areas of the right triangle (120) and the equilateral triangle (292.7), and add them together.
Exam Tip: If the value of \( \sqrt{3} \) is not specified, you may write your final answer as \( 120 + 169\sqrt{3} \text{ cm}^2 \).

Question 33. Mayank made a picture of an aeroplane with paper as shown in figure calculate total area of paper used.
Answer: The total area of the paper aeroplane is calculated by summing the areas of its five distinct sections:
Part I (Top isosceles triangle): Sides are \( 5 \text{ cm} \), \( 5 \text{ cm} \), and base \( 1 \text{ cm} \).
\( s = \frac{5 + 5 + 1}{2} = 5.5 \text{ cm} \) \[ \text{Area I} = \sqrt{5.5(5.5-5)(5.5-5)(5.5-1)} = \sqrt{5.5 \times 0.5 \times 0.5 \times 4.5} \approx 2.49 \text{ cm}^2 \]
Part II (Middle rectangle): Length \( 5.5 \text{ cm} \) and width \( 1 \text{ cm} \). \[ \text{Area II} = \text{length} \times \text{width} = 5.5 \times 1 = 5.5 \text{ cm}^2 \]
Part III (Bottom trapezium): Parallel sides are \( 1 \text{ cm} \) and \( 2 \text{ cm} \), non-parallel sides are \( 1 \text{ cm} \).
The height is \( h = \sqrt{1^2 - (0.5)^2} = \sqrt{0.75} \approx 0.866 \text{ cm} \). \[ \text{Area III} = \frac{1}{2} \times (1 + 2) \times 0.866 \approx 1.3 \text{ cm}^2 \]
Parts IV and V (Two triangular wings): Each is a right triangle with base \( 1.5 \text{ cm} \) and height \( 6 \text{ cm} \). \[ \text{Area IV + V} = 2 \times \left(\frac{1}{2} \times 1.5 \times 6\right) = 9 \text{ cm}^2 \]
The total area of paper used is: \[ \text{Total Area} = \text{Area I} + \text{Area II} + \text{Area III} + \text{Area IV + V} \]
\( \implies \text{Total Area} = 2.49 + 5.5 + 1.3 + 9 = 18.29 \text{ cm}^2 \) In simple words: Break the aeroplane down into five pieces (triangle, rectangle, trapezium, and two wings). Calculate each area separately and sum them to get 18.29.
Exam Tip: Be methodical and write down calculations for each of the five parts separately to avoid arithmetic errors.

Question 34. A kite in the shape of a square with diagonal 32 cm and an isosceles triangle of base 8 cm and equal sides are 6cm how much paper is required to build the kite.
Answer: To find the amount of paper needed, we calculate the area of both the square part and the isosceles triangle part of the kite.

1. Area of the square part:
The diagonal of the square, \( d = 32 \text{ cm} \).
The area of a square using its diagonal is:
\[ \text{Area of square} = \frac{1}{2} \times d^2 \]
\[ \text{Area of square} = \frac{1}{2} \times 32 \times 32 = 512 \text{ cm}^2 \]

2. Area of the isosceles triangle part:
The base of the triangle, \( b = 8 \text{ cm} \).
The equal sides, \( a = 6 \text{ cm} \).
First, find the height (\( h \)) of the triangle using Pythagoras' theorem:
\[ h = \sqrt{a^2 - \left(\frac{b}{2}\right)^2} \]
\[ h = \sqrt{6^2 - 4^2} = \sqrt{36 - 16} = \sqrt{20} = 2\sqrt{5} \approx 4.47 \text{ cm} \]
The area of the triangle is:
\[ \text{Area of triangle} = \frac{1}{2} \times \text{base} \times \text{height} \]
\[ \text{Area of triangle} = \frac{1}{2} \times 8 \times 2\sqrt{5} = 8\sqrt{5} \approx 17.89 \text{ cm}^2 \]

3. Total paper required:
\[ \text{Total Area} = \text{Area of square} + \text{Area of triangle} \]
\[ \text{Total Area} = 512 + 8\sqrt{5} \approx 512 + 17.89 = 529.89 \text{ cm}^2 \]

In simple words: Add the area of the square to the area of the small triangle at the bottom to find the total paper. The square area is 512 sq. cm, and the triangle area is around 17.89 sq. cm, making the total about 529.89 sq. cm.

Exam Tip: Remember that the area of a square is half the square of its diagonal. Using this formula directly saves time compared to finding the side length first.

Question 35. The perimeter of a right triangle is 60 cm and its hypotenuse is 26 cm. Find area of triangle and its other two sides.
Answer: Let the two perpendicular sides of the right-angled triangle be \( a \) and \( b \), and the hypotenuse be \( c = 26 \text{ cm} \).
The perimeter of the triangle is:
\( a + b + c = 60 \)
\( \implies a + b + 26 = 60 \)
\( \implies a + b = 34 \)
Using the Pythagorean identity:
\( a^2 + b^2 = c^2 \)
\( \implies a^2 + b^2 = 26^2 = 676 \)
We can find the product \( ab \) using the algebraic expansion:
\[ (a + b)^2 = a^2 + b^2 + 2ab \]
\[ 34^2 = 676 + 2ab \]
\[ 1156 = 676 + 2ab \]
\( \implies 2ab = 480 \)
\( \implies ab = 240 \)
The area of a right-angled triangle is:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times ab = \frac{1}{2} \times 240 = 120 \text{ cm}^2 \]
To find the individual sides, we calculate:
\[ (a - b)^2 = (a + b)^2 - 4ab = 34^2 - 4(240) = 1156 - 960 = 196 \]
\( \implies a - b = \sqrt{196} = 14 \)
Now we solve the simultaneous equations:
\( a + b = 34 \)
\( a - b = 14 \)
Adding these equations:
\( 2a = 48 \)
\( \implies a = 24 \text{ cm} \)
Then:
\[ b = 34 - 24 = 10 \text{ cm} \]
Thus, the other two sides are 24 cm and 10 cm, and the area is 120 cm^2.
In simple words: Use the perimeter and hypotenuse to find the sum of the other two sides, which is 34 cm. Applying the right-triangle rule shows the two sides are 24 cm and 10 cm, giving an area of 120 sq. cm.

Exam Tip: For right triangles, look for Pythagorean triplets. Since 5, 12, 13 is a common triplet, doubling it gives 10, 24, 26, which fits this problem perfectly and can help you verify your solution quickly.

Question 36. A trapezium PBCQ with its parallel sides QC and PB in the ratio 7:5 is cut from a rectangle ABCD, if area of trapezium is 4/7 part of the area of rectangle, find the length of QC and PB.
Answer: From the rectangle ABCD, we know:
- The base \( DC = 21 \text{ cm} \).
- The height \( BC = 5 \text{ cm} \).
The area of the rectangle ABCD is:
\[ \text{Area of rectangle} = \text{base} \times \text{height} = 21 \times 5 = 105 \text{ cm}^2 \]
Let the parallel sides of the trapezium PBCQ be \( QC = 7x \) and \( PB = 5x \).
The height of the trapezium is equal to the width of the rectangle, which is \( BC = 5 \text{ cm} \).
The area of the trapezium PBCQ is:
\[ \text{Area of trapezium} = \frac{1}{2} \times (\text{Sum of parallel sides}) \times \text{height} \]
\[ \text{Area of trapezium} = \frac{1}{2} \times (7x + 5x) \times 5 = \frac{1}{2} \times 12x \times 5 = 30x \]
We are given that the area of the trapezium is \( \frac{4}{7} \) of the area of the rectangle:
\[ 30x = \frac{4}{7} \times 105 \]
\[ 30x = 4 \times 15 = 60 \]
\( \implies x = 2 \)
Now, calculate the lengths:
\[ QC = 7 \times 2 = 14 \text{ cm} \]
\[ PB = 5 \times 2 = 10 \text{ cm} \]
In simple words: The total area of the rectangle is 105 sq. cm, so the trapezium's area is 60 sq. cm. Using the area of a trapezium formula with a height of 5 cm, we find the sides are 14 cm and 10 cm.

Exam Tip: Always identify that the height of the trapezium cut from the rectangle is the same as the rectangle's breadth. This is the key link to solving the problem.

Question 37. A trapezium whose parallel sides are 25 cm and 10 cm. The non-parallel sides are 14 cm and 13 cm. find the area of the trapezium .
Answer: Let ABCD be the trapezium where parallel sides are \( AB = 25 \text{ cm} \) and \( CD = 10 \text{ cm} \), and non-parallel sides are \( AD = 14 \text{ cm} \) and \( BC = 13 \text{ cm} \).
Draw \( DE \parallel BC \) such that \( E \) lies on \( AB \).
Thus, DEBC is a parallelogram where:
\[ DE = BC = 13 \text{ cm} \]
\[ EB = CD = 10 \text{ cm} \]
The remaining segment of the base is:
\[ AE = AB - EB = 25 - 10 = 15 \text{ cm} \]
Now, consider triangle ADE with side lengths \( 14 \text{ cm} \), \( 13 \text{ cm} \), and \( 15 \text{ cm} \).
Calculate the semi-perimeter of triangle ADE:
\[ s = \frac{14 + 13 + 15}{2} = 21 \text{ cm} \]
Using Heron's Formula for the area of triangle ADE:
\[ \text{Area} = \sqrt{s(s - AD)(s - DE)(s - AE)} \]
\[ \text{Area} = \sqrt{21(21 - 14)(21 - 13)(21 - 15)} \]
\[ \text{Area} = \sqrt{21 \times 7 \times 8 \times 6} = \sqrt{7056} = 84 \text{ cm}^2 \]
Also, the area of triangle ADE can be written as:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]
\[ 84 = \frac{1}{2} \times 15 \times h \]
\( \implies h = \frac{168}{15} = 11.2 \text{ cm} \)
This height \( h \) is also the height of the trapezium.
The area of the trapezium ABCD is:
\[ \text{Area of trapezium} = \frac{1}{2} \times (\text{Sum of parallel sides}) \times \text{height} \]
\[ \text{Area of trapezium} = \frac{1}{2} \times (25 + 10) \times 11.2 = 35 \times 5.6 = 196 \text{ cm}^2 \]
In simple words: Divide the trapezium into a parallelogram and a triangle. Use Heron's formula to find the triangle's area, which gives the height as 11.2 cm, and then calculate the total trapezium area as 196 sq. cm.

Exam Tip: When solving trapezium problems with all four sides given, drawing a line parallel to one of the non-parallel sides is the standard method to split it into a parallelogram and a triangle.

Question 38. The sides of a quadrilateral ABCD, taken in order are 5 cm,12 cm,14 cm and 15 cm respectively, and angle contained between first two sides is a right angle. Find its area.
Answer: Let the quadrilateral be ABCD with side lengths:
\[ AB = 5 \text{ cm}, \quad BC = 12 \text{ cm}, \quad CD = 14 \text{ cm}, \quad DA = 15 \text{ cm} \]
The angle between the first two sides (AB and BC) is \( \angle B = 90^\circ \).
Draw the diagonal AC to split the quadrilateral into two triangles: \( \Delta ABC \) and \( \Delta ADC \).

1. Area of right-angled triangle ABC:
\[ \text{Area of } \Delta ABC = \frac{1}{2} \times \text{base} \times \text{height} \]
\[ \text{Area of } \Delta ABC = \frac{1}{2} \times 5 \times 12 = 30 \text{ cm}^2 \]
Using Pythagoras' theorem to find diagonal AC:
\[ AC = \sqrt{AB^2 + BC^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \text{ cm} \]

2. Area of triangle ADC:
The sides of \( \Delta ADC \) are \( AD = 15 \text{ cm} \), \( CD = 14 \text{ cm} \), and \( AC = 13 \text{ cm} \).
Calculate the semi-perimeter:
\[ s = \frac{15 + 14 + 13}{2} = 21 \text{ cm} \]
Using Heron's Formula for the area of \( \Delta ADC \):
\[ \text{Area of } \Delta ADC = \sqrt{s(s - AD)(s - CD)(s - AC)} \]
\[ \text{Area of } \Delta ADC = \sqrt{21(21 - 15)(21 - 14)(21 - 13)} \]
\[ \text{Area of } \Delta ADC = \sqrt{21 \times 6 \times 7 \times 8} = \sqrt{7056} = 84 \text{ cm}^2 \]

3. Total Area of Quadrilateral ABCD:
\[ \text{Total Area} = \text{Area of } \Delta ABC + \text{Area of } \Delta ADC = 30 + 84 = 114 \text{ cm}^2 \]
In simple words: Split the quadrilateral into two triangles by drawing a diagonal. The right triangle has an area of 30 sq. cm, and the other triangle has an area of 84 sq. cm, giving a total of 114 sq. cm.

Exam Tip: Remember to state that you are drawing diagonal AC to form two triangles. Breaking down quadrilaterals into triangles is the standard way to apply Heron's formula.

Question 39. A rhombus sheet, whose perimeter is 32 cm and whose one diagonal is 10 cm long, is painted on both sides at the rate of 5 per sq. cm. Find the cost of painting.
Answer: Let ABCD be the rhombus sheet. The perimeter of the rhombus is \( 32 \text{ cm} \). Since all four sides of a rhombus are equal:
\[ \text{Side length, } a = \frac{32}{4} = 8 \text{ cm} \]
Let one diagonal be \( d_1 = 10 \text{ cm} \). A diagonal divides the rhombus into two congruent triangles, each having side lengths \( 8 \text{ cm} \), \( 8 \text{ cm} \), and \( 10 \text{ cm} \).
Using Heron's Formula to find the area of one such triangle:
\[ s = \frac{8 + 8 + 10}{2} = 13 \text{ cm} \]
\[ \text{Area of one triangle} = \sqrt{13(13 - 8)(13 - 8)(13 - 10)} \]
\[ \text{Area of one triangle} = \sqrt{13 \times 5 \times 5 \times 3} = 5\sqrt{39} \approx 5 \times 6.245 = 31.225 \text{ cm}^2 \]
The area of the entire rhombus is:
\[ \text{Area of rhombus} = 2 \times 5\sqrt{39} = 10\sqrt{39} \approx 62.45 \text{ cm}^2 \]
Since the sheet is painted on both sides, the total area to be painted is:
\[ \text{Total painted area} = 2 \times 10\sqrt{39} = 20\sqrt{39} \approx 124.90 \text{ cm}^2 \]
The rate of painting is Rs. 5 per sq. cm.
The total cost of painting is:
\[ \text{Cost} = \text{Total Area} \times \text{Rate} \]
\[ \text{Cost} = 20\sqrt{39} \times 5 = 100\sqrt{39} \approx 100 \times 6.245 = \text{Rs. } 624.50 \]
In simple words: Find the area of one face of the rhombus sheet, which is around 62.45 sq. cm. Since it is painted on both sides, double this area and multiply by the rate of Rs. 5 to get a total cost of Rs. 624.50.

Exam Tip: Pay close attention to "painted on both sides" in the question statement. Forgetting to double the area for double-sided painting is a very common error.

Question 40. The perimeter of a right triangle is 60 cm and its hypotenuse is 26 cm. Find other two sides and area of triangle?
Answer: Let the two legs of the right triangle be \( x \) and \( y \), and the hypotenuse be \( 26 \text{ cm} \).
The perimeter is given as:
\( x + y + 26 = 60 \)
\( \implies x + y = 34 \)
Using the Pythagorean relation:
\( x^2 + y^2 = 26^2 = 676 \)
We know that:
\[ (x + y)^2 = x^2 + y^2 + 2xy \]
Substitute the values:
\[ 34^2 = 676 + 2xy \]
\[ 1156 = 676 + 2xy \]
\( \implies 2xy = 480 \)
\( \implies xy = 240 \)
The area of the right-angled triangle is:
\[ \text{Area} = \frac{1}{2} \times xy = \frac{1}{2} \times 240 = 120 \text{ cm}^2 \]
To determine the side lengths \( x \) and \( y \), we calculate:
\[ (x - y)^2 = (x + y)^2 - 4xy = 34^2 - 4(240) = 1156 - 960 = 196 \]
\( \implies x - y = \sqrt{196} = 14 \)
We have two linear equations:
\( x + y = 34 \)
\( x - y = 14 \)
Adding them together:
\( 2x = 48 \)
\( \implies x = 24 \text{ cm} \)
Subtracting them:
\( 2y = 20 \)
\( \implies y = 10 \text{ cm} \)
Thus, the other two sides are 24 cm and 10 cm, and the area of the triangle is 120 cm^2.
In simple words: Find the sum of the legs, which is 34 cm. Combining this with the hypotenuse using right-angle math shows the sides are 24 cm and 10 cm, and the area is 120 sq. cm.

Exam Tip: Always double check that your calculated side lengths satisfy the Pythagorean theorem: \( 10^2 + 24^2 = 100 + 576 = 676 = 26^2 \). This verifies your algebra is completely correct.

Question 41. A field is in the shape of trapezium whose parallel sides are 25m and 10m. The non-parallel sides are 14m and 13m. Find the area of the field.
Answer: Let ABCD be the trapezium-shaped field, where parallel sides are \( AB = 25 \text{ m} \) and \( CD = 10 \text{ m} \), and the non-parallel sides are \( AD = 14 \text{ m} \) and \( BC = 13 \text{ m} \).
Draw \( DE \parallel BC \) such that \( E \) lies on \( AB \).
This forms a parallelogram DEBC and a triangle ADE:
\[ DE = BC = 13 \text{ m} \]
\[ EB = CD = 10 \text{ m} \]
The base of the triangle ADE is:
\[ AE = AB - EB = 25 - 10 = 15 \text{ m} \]
For triangle ADE with side lengths \( 14 \text{ m} \), \( 13 \text{ m} \), and \( 15 \text{ m} \), the semi-perimeter is:
\[ s = \frac{14 + 13 + 15}{2} = 21 \text{ m} \]
Using Heron's Formula for the area of triangle ADE:
\[ \text{Area of } \Delta ADE = \sqrt{s(s - AD)(s - DE)(s - AE)} \]
\[ \text{Area of } \Delta ADE = \sqrt{21(21 - 14)(21 - 13)(21 - 15)} \]
\[ \text{Area of } \Delta ADE = \sqrt{21 \times 7 \times 8 \times 6} = \sqrt{7056} = 84 \text{ m}^2 \]
Using the triangle area formula to find the height:
\[ \text{Area of } \Delta ADE = \frac{1}{2} \times \text{base} \times \text{height} \]
\[ 84 = \frac{1}{2} \times 15 \times h \]
\( \implies h = \frac{168}{15} = 11.2 \text{ m} \)
The area of the trapezium field is:
\[ \text{Area of field} = \frac{1}{2} \times (\text{Sum of parallel sides}) \times \text{height} \]
\[ \text{Area of field} = \frac{1}{2} \times (25 + 10) \times 11.2 = 35 \times 5.6 = 196 \text{ m}^2 \]
In simple words: Split the field into a triangle and a parallelogram. Use Heron's formula to find the triangle's area, which allows you to find the height of 11.2 m. Then, use the trapezium formula to find the total area of 196 sq. m.

Exam Tip: Ensure that you clearly state the construction step of drawing DE parallel to BC. Writing down the steps of construction is important for securing full marks in descriptive geometry questions.

Most Important Questions

Question 1. What is the area of an equilateral triangle of side 2a.
Answer: The formula for the area of an equilateral triangle with side length \( s \) is:
\[ \text{Area} = \frac{\sqrt{3}}{4} s^2 \]
Given that the side length \( s = 2a \), substitute this into the formula:
\[ \text{Area} = \frac{\sqrt{3}}{4} (2a)^2 \]
\[ \text{Area} = \frac{\sqrt{3}}{4} \times 4a^2 = \sqrt{3}a^2 \]
In simple words: The area of an equilateral triangle is found by multiplying the square of its side by root three over four. For a side of 2a, this simplifies to root three times a squared.

Exam Tip: Do not forget to square the entire term \( 2a \) to get \( 4a^2 \). A common mistake is writing \( 2a^2 \) instead of \( 4a^2 \).

Question 2. What will be the area of a right angled triangle whose base is 12 cm and hypotenuse is 13 cm.
Answer: Let the altitude (height) of the right-angled triangle be \( h \).
By Pythagoras' theorem:
\[ \text{base}^2 + \text{height}^2 = \text{hypotenuse}^2 \]
\[ 12^2 + h^2 = 13^2 \]
\[ 144 + h^2 = 169 \]
\( \implies h^2 = 25 \)
\( \implies h = 5 \text{ cm} \)
Now, calculate the area of the triangle:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]
\[ \text{Area} = \frac{1}{2} \times 12 \times 5 = 30 \text{ cm}^2 \]
In simple words: Use Pythagoras' theorem to find the missing height, which is 5 cm. Then multiply half of the base by the height to get the area of 30 sq. cm.

Exam Tip: Memorizing common Pythagorean triples such as (5, 12, 13) allows you to find the third side instantly and spend more time on calculations.

Question 3. Find the area of a triangle whose sides are respectively 150 cm, 120 cm and 200 cm.
Answer: Let the side lengths of the triangle be \( a = 150 \text{ cm} \), \( b = 120 \text{ cm} \), and \( c = 200 \text{ cm} \).
First, find the semi-perimeter \( s \):
\[ s = \frac{a + b + c}{2} = \frac{150 + 120 + 200}{2} = \frac{470}{2} = 235 \text{ cm} \]
Now, use Heron's Formula to calculate the area:
\[ \text{Area} = \sqrt{s(s - a)(s - b)(s - c)} \]
\[ \text{Area} = \sqrt{235(235 - 150)(235 - 120)(235 - 200)} \]
\[ \text{Area} = \sqrt{235 \times 85 \times 115 \times 35} \]
Break down into prime factors:
\[ 235 = 5 \times 47 \]
\[ 85 = 5 \times 17 \]
\[ 115 = 5 \times 23 \]
\[ 35 = 5 \times 7 \]
Substitute the factors:
\[ \text{Area} = \sqrt{(5 \times 47) \times (5 \times 17) \times (5 \times 23) \times (5 \times 7)} \]
\[ \text{Area} = \sqrt{5^4 \times 47 \times 17 \times 23 \times 7} \]
\[ \text{Area} = 25\sqrt{128639} \approx 25 \times 358.66 = 8966.56 \text{ cm}^2 \]
In simple words: Find the semi-perimeter, which is 235 cm. Apply Heron's formula by multiplying 235 by the differences of each side, then take the square root to get an area of about 8966.56 sq. cm.

Exam Tip: When dealing with large numbers under the square root, factorize them instead of multiplying them out directly. This keeps the math simpler and reduces errors.

Question 4. The perimeter of a triangular field is 540 m and its sides are in the ratio 25:17:12. Find the area of the triangle.
Answer: Let the sides of the triangular field be \( 25x \), \( 17x \), and \( 12x \).
The perimeter of the field is \( 540 \text{ m} \):
\[ 25x + 17x + 12x = 540 \]
\[ 54x = 540 \]
\( \implies x = 10 \)
Thus, the sides are:
\[ a = 25 \times 10 = 250 \text{ m} \]
\[ b = 17 \times 10 = 170 \text{ m} \]
\[ c = 12 \times 10 = 120 \text{ m} \]
The semi-perimeter of the field is:
\[ s = \frac{540}{2} = 270 \text{ m} \]
Using Heron's Formula:
\[ \text{Area} = \sqrt{s(s - a)(s - b)(s - c)} \]
\[ \text{Area} = \sqrt{270(270 - 250)(270 - 170)(270 - 120)} \]
\[ \text{Area} = \sqrt{270 \times 20 \times 100 \times 150} \]
\[ \text{Area} = \sqrt{81,000,000} = 9000 \text{ m}^2 \]
In simple words: Find the actual side lengths, which are 250 m, 170 m, and 120 m. Then use Heron's formula to get the area of 9000 sq. m.

Exam Tip: Write down the ratio equations clearly. Finding the multiplier \( x = 10 \) first is the standard method for ratio-based triangle problems.

Question 5. A triangle and a parallelogram are on the same base and have the same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm, and the parallelogram stands on the base 28 cm, find the height of the parallelogram.
Answer: First, find the area of the triangle with sides \( a = 26 \text{ cm} \), \( b = 28 \text{ cm} \), and \( c = 30 \text{ cm} \).
The semi-perimeter is:
\[ s = \frac{26 + 28 + 30}{2} = \frac{84}{2} = 42 \text{ cm} \]
Using Heron's Formula:
\[ \text{Area} = \sqrt{s(s - a)(s - b)(s - c)} \]
\[ \text{Area} = \sqrt{42(42 - 26)(42 - 28)(42 - 30)} \]
\[ \text{Area} = \sqrt{42 \times 16 \times 14 \times 12} \]
Break down into factors to simplify:
\[ \text{Area} = \sqrt{(14 \times 3) \times 16 \times 14 \times (3 \times 4)} \]
\[ \text{Area} = \sqrt{14^2 \times 3^2 \times 16 \times 4} = 14 \times 3 \times 4 \times 2 = 336 \text{ cm}^2 \]
Since the triangle and the parallelogram have the same area and the same base \( 28 \text{ cm} \):
\[ \text{Area of parallelogram} = \text{base} \times \text{height} \]
\[ 336 = 28 \times h \]
\( \implies h = \frac{336}{28} = 12 \text{ cm} \)
In simple words: Find the area of the triangle, which is 336 sq. cm. Since the parallelogram has the same area and stands on a 28 cm base, its height is 12 cm.

Exam Tip: Keep in mind that the area formula for a parallelogram is just base times height, unlike triangles which require the half factor.

Question 6. The perimeter of a triangular park is 240 m. If two of its sides are 78 m and 50 m, find the length of the perpendicular on the side of length 50 m from the opposite vertex.
Answer: Let the sides of the triangular park be \( a = 78 \text{ m} \), \( b = 50 \text{ m} \), and \( c \).
The perimeter is \( 240 \text{ m} \):
\( a + b + c = 240 \)
\( \implies 78 + 50 + c = 240 \)
\( \implies 128 + c = 240 \)
\( \implies c = 112 \text{ m} \)
The semi-perimeter is:
\[ s = \frac{240}{2} = 120 \text{ m} \]
Using Heron's Formula to find the area:
\[ \text{Area} = \sqrt{s(s - a)(s - b)(s - c)} \]
\[ \text{Area} = \sqrt{120(120 - 78)(120 - 50)(120 - 112)} \]
\[ \text{Area} = \sqrt{120 \times 42 \times 70 \times 8} \]
\[ \text{Area} = \sqrt{2,822,400} = 1680 \text{ m}^2 \]
The length of the perpendicular on the side of \( 50 \text{ m} \) (which acts as the base) is:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]
\[ 1680 = \frac{1}{2} \times 50 \times h \]
\[ 1680 = 25 \times h \]
\( \implies h = \frac{1680}{25} = 67.2 \text{ m} \)
In simple words: Find the third side first, which is 112 m. Use Heron's formula to get the area of 1680 sq. m, and then divide by half of the base (25) to find the height of 67.2 m.

Exam Tip: The "perpendicular on the side of length 50 m from the opposite vertex" is simply the altitude corresponding to that side. Treat 50 m as the base in your second area formula.

Question 7. Find the percentage increase in the area of a triangle if its each side is doubled.
Answer: Let the original sides of the triangle be \( a \), \( b \), and \( c \).
The original semi-perimeter is \( s = \frac{a + b + c}{2} \).
The original area \( A \) is:
\[ A = \sqrt{s(s - a)(s - b)(s - c)} \]
If each side is doubled, the new sides are \( 2a \), \( 2b \), and \( 2c \).
The new semi-perimeter \( s' \) is:
\[ s' = \frac{2a + 2b + 2c}{2} = 2s \]
The new area \( A' \) is:
\[ A' = \sqrt{s'(s' - 2a)(s' - 2b)(s' - 2c)} \]
\[ A' = \sqrt{2s(2s - 2a)(2s - 2b)(2s - 2c)} \]
\[ A' = \sqrt{16s(s - a)(s - b)(s - c)} \]
\[ A' = 4\sqrt{s(s - a)(s - b)(s - c)} = 4A \]
The increase in area is:
\[ \text{Increase} = A' - A = 4A - A = 3A \]
The percentage increase is:
\[ \text{Percentage Increase} = \frac{\text{Increase}}{\text{Original Area}} \times 100\% \]
\[ \text{Percentage Increase} = \frac{3A}{A} \times 100\% = 300\% \]
In simple words: When you double every side of a triangle, its area becomes four times larger. This means the area has increased by three times its original size, which is a 300% increase.

Exam Tip: Be careful with the distinction between "becomes 4 times" and "increases by". The percentage increase is calculated on the change, which is \( 4A - A = 3A \), yielding 300%.

Question 8. A skirt is made by stitching 10 triangular pieces of cloth of two different colors, each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each color is required for the skirt.
Answer: The skirt consists of 10 triangular pieces of two different colors, which means there are 5 pieces of each color.
First, find the area of one triangular piece with sides \( 20 \text{ cm} \), \( 50 \text{ cm} \), and \( 50 \text{ cm} \).
The semi-perimeter of this triangle is:
\[ s = \frac{20 + 50 + 50}{2} = 60 \text{ cm} \]
Using Heron's Formula:
\[ \text{Area of one piece} = \sqrt{s(s - 20)(s - 50)(s - 50)} \]
\[ \text{Area of one piece} = \sqrt{60 \times 40 \times 10 \times 10} \]
\[ \text{Area of one piece} = \sqrt{240,000} = 200\sqrt{6} \text{ cm}^2 \]
Since there are 5 pieces of each color, the area of cloth required for each color is:
\[ \text{Total area for one color} = 5 \times 200\sqrt{6} = 1000\sqrt{6} \text{ cm}^2 \]
Using \( \sqrt{6} \approx 2.449 \):
\[ \text{Area} \approx 1000 \times 2.449 = 2449 \text{ cm}^2 \]
In simple words: Calculate the area of one triangle, which is 200 times the square root of 6 sq. cm. Since there are 5 triangles of each color, multiply this by 5 to get 1000 root 6 (or about 2449) sq. cm.

Exam Tip: Make sure to divide the total number of pieces (10) by 2, as the pieces are of two different colors. A common mistake is multiplying by 10 instead of 5.

Question 9. Find the area of the quadrilateral ABCD in which AD = 24 cm, BAD = 90° and BCD forms an equilateral triangle whose each side is equal to 26 cm.
Answer: The quadrilateral ABCD is split into two triangles by diagonal BD: right-angled triangle ABD and equilateral triangle BCD.
Since BCD is an equilateral triangle with side lengths of \( 26 \text{ cm} \), the diagonal \( BD = 26 \text{ cm} \).

1. Area of right-angled triangle ABD:
In \( \Delta ABD \), \( \angle BAD = 90^\circ \), \( AD = 24 \text{ cm} \), and the hypotenuse \( BD = 26 \text{ cm} \).
By Pythagoras' theorem:
\[ AB = \sqrt{BD^2 - AD^2} = \sqrt{26^2 - 24^2} = \sqrt{676 - 576} = \sqrt{100} = 10 \text{ cm} \]
The area of \( \Delta ABD \) is:
\[ \text{Area of } \Delta ABD = \frac{1}{2} \times AB \times AD = \frac{1}{2} \times 10 \times 24 = 120 \text{ cm}^2 \]

2. Area of equilateral triangle BCD:
The side of the equilateral triangle is \( s = 26 \text{ cm} \).
The area of \( \Delta BCD \) is:
\[ \text{Area of } \Delta BCD = \frac{\sqrt{3}}{4} s^2 = \frac{\sqrt{3}}{4} \times 26^2 = \frac{\sqrt{3}}{4} \times 676 = 169\sqrt{3} \text{ cm}^2 \]
Using \( \sqrt{3} \approx 1.732 \):
\[ \text{Area of } \Delta BCD \approx 169 \times 1.732 = 292.71 \text{ cm}^2 \]

3. Total Area of Quadrilateral ABCD:
\[ \text{Total Area} = \text{Area of } \Delta ABD + \text{Area of } \Delta BCD \]
\[ \text{Total Area} = 120 + 169\sqrt{3} \approx 120 + 292.71 = 412.71 \text{ cm}^2 \]
In simple words: Find the side AB using Pythagoras' theorem to get 10 cm, then calculate the right triangle area as 120 sq. cm. Add this to the area of the equilateral triangle (approx. 292.71 sq. cm) to get a total of 412.71 sq. cm.

Exam Tip: Clearly label which triangle is right-angled and which is equilateral before applying their respective area formulas.

Question 10. Find the area of a rhombus whose perimeter is 80 cm and one of its diagonal is 24 cm.
Answer: Let ABCD be the rhombus. The perimeter is \( 80 \text{ cm} \). Since all four sides of a rhombus are equal:
\[ \text{Side length, } a = \frac{80}{4} = 20 \text{ cm} \]
Let one diagonal be \( d_1 = 24 \text{ cm} \). The diagonals of a rhombus bisect each other at right angles. Let the other diagonal be \( d_2 \).
Using Pythagoras' theorem on one of the four small right triangles formed inside:
\[ \left(\frac{d_1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2 = a^2 \]
\[ 12^2 + \left(\frac{d_2}{2}\right)^2 = 20^2 \]
\[ 144 + \left(\frac{d_2}{2}\right)^2 = 400 \]
\( \implies \left(\frac{d_2}{2}\right)^2 = 256 \)
\( \implies \frac{d_2}{2} = 16 \)
\( \implies d_2 = 32 \text{ cm} \)
The area of the rhombus is:
\[ \text{Area} = \frac{1}{2} \times d_1 \times d_2 \]
\[ \text{Area} = \frac{1}{2} \times 24 \times 32 = 384 \text{ cm}^2 \]
In simple words: Divide the perimeter by 4 to find the side length of 20 cm. Use right-triangle rules to find the half of the other diagonal, which is 16 cm, meaning the full second diagonal is 32 cm. Finally, use the area formula to get 384 sq. cm.

Exam Tip: Alternatively, you can divide the rhombus into two triangles of sides 20 cm, 20 cm, and 24 cm, and use Heron's formula. Both methods yield the exact same result.

Question 11. Two parallel sides of a trapezium are 60 cm and 77 cm and the others sides are 25 cm and 26 cm. Find the area of the trapezium.
Answer: Let ABCD be the trapezium with parallel sides \( AB = 77 \text{ cm} \) and \( CD = 60 \text{ cm} \), and non-parallel sides \( AD = 25 \text{ cm} \) and \( BC = 26 \text{ cm} \).
Draw \( DE \parallel BC \) such that \( E \) lies on \( AB \).
This splits the trapezium into a parallelogram DEBC and a triangle ADE:
\[ DE = BC = 26 \text{ cm} \]
\[ EB = CD = 60 \text{ cm} \]
The base of triangle ADE is:
\[ AE = AB - EB = 77 - 60 = 17 \text{ cm} \]
In triangle ADE, the sides are \( AD = 25 \text{ cm} \), \( DE = 26 \text{ cm} \), and \( AE = 17 \text{ cm} \).
The semi-perimeter is:
\[ s = \frac{25 + 26 + 17}{2} = 34 \text{ cm} \]
Using Heron's Formula:
\[ \text{Area of } \Delta ADE = \sqrt{s(s - AD)(s - DE)(s - AE)} \]
\[ \text{Area of } \Delta ADE = \sqrt{34(34 - 25)(34 - 26)(34 - 17)} \]
\[ \text{Area of } \Delta ADE = \sqrt{34 \times 9 \times 8 \times 17} = 204 \text{ cm}^2 \]
Find the altitude \( h \) of triangle ADE (which is also the height of the trapezium):
\[ \text{Area of } \Delta ADE = \frac{1}{2} \times \text{base} \times \text{height} \]
\[ 204 = \frac{1}{2} \times 17 \times h \]
\( \implies h = \frac{408}{17} = 24 \text{ cm} \)
The area of the trapezium ABCD is:
\[ \text{Area} = \frac{1}{2} \times (\text{Sum of parallel sides}) \times \text{height} \]
\[ \text{Area} = \frac{1}{2} \times (77 + 60) \times 24 = 137 \times 12 = 1644 \text{ cm}^2 \]
In simple words: Divide the trapezium into a triangle and a parallelogram. Use Heron's formula on the triangle to get its area as 204 sq. cm, which gives a height of 24 cm. Then calculate the total area as 1644 sq. cm.

Exam Tip: Drawing a line parallel to one of the non-parallel sides is a highly effective construction that works for any trapezium where all side lengths are known.

Question 12. Radha has a piece of land which is in the shape of a rhombus. She wants her son and daughter to cultivate equal areas of land, thus she divides the area of the rhombus equally amongst the two. If the perimeter of land is 400 m and of the diagonals is 160 m, How much area will each one get?
Answer: Let ABCD be the rhombus-shaped land. The perimeter of the rhombus is \( 400 \text{ m} \). Since all four sides are equal, the side length is:
\[ \text{Side, } a = \frac{400}{4} = 100 \text{ m} \]
One diagonal is \( d_1 = 160 \text{ m} \). The diagonals of a rhombus bisect each other at right angles, so half of the first diagonal is \( 80 \text{ m} \).
Let the other diagonal be \( d_2 \). Using Pythagoras' theorem on one of the inner right triangles:
\[ 80^2 + \left(\frac{d_2}{2}\right)^2 = 100^2 \]
\[ 6400 + \left(\frac{d_2}{2}\right)^2 = 10000 \]
\( \implies \left(\frac{d_2}{2}\right)^2 = 3600 \)
\( \implies \frac{d_2}{2} = 60 \)
\( \implies d_2 = 120 \text{ m} \)
The total area of the rhombus land is:
\[ \text{Total Area} = \frac{1}{2} \times d_1 \times d_2 = \frac{1}{2} \times 160 \times 120 = 9600 \text{ m}^2 \]
Since the land is divided equally between her son and daughter:
\[ \text{Area each child gets} = \frac{9600}{2} = 4800 \text{ m}^2 \]
In simple words: Calculate the total area of the rhombus land first, which is 9600 sq. m. Since it is shared equally between the son and daughter, divide this by two to get 4800 sq. m for each.

Exam Tip: Always read carefully what is being asked. The question asks how much area each child will get, so dividing the total area by 2 is a crucial final step.

Question 13. A floral design on the wall is made up of 16 tiles which are triangular, the sides are of the triangle are 9 cm, 28 cm and 35 cm. Find the cost of polishing the tiles at the rate of 50 paise per cm2.
Answer: There are 16 triangular tiles in total. First, find the area of a single triangular tile with sides \( a = 9 \text{ cm} \), \( b = 28 \text{ cm} \), and \( c = 35 \text{ cm} \).
The semi-perimeter is:
\[ s = \frac{9 + 28 + 35}{2} = \frac{72}{2} = 36 \text{ cm} \]
Using Heron's Formula:
\[ \text{Area of one tile} = \sqrt{s(s - a)(s - b)(s - c)} \]
\[ \text{Area of one tile} = \sqrt{36(36 - 9)(36 - 28)(36 - 35)} \]
\[ \text{Area of one tile} = \sqrt{36 \times 27 \times 8 \times 1} \]
\[ \text{Area of one tile} = \sqrt{36 \times (9 \times 3) \times (4 \times 2)} \]
\[ \text{Area of one tile} = 6 \times 3 \times 2 \times \sqrt{6} = 36\sqrt{6} \text{ cm}^2 \]
The total area of all 16 tiles is:
\[ \text{Total Area} = 16 \times 36\sqrt{6} = 576\sqrt{6} \text{ cm}^2 \]
Using \( \sqrt{6} \approx 2.45 \):
\[ \text{Total Area} \approx 576 \times 2.45 = 1411.2 \text{ cm}^2 \]
The rate of polishing is 50 paise per sq. cm, which is Rs. 0.50 per sq. cm.
The total cost of polishing is:
\[ \text{Cost} = \text{Total Area} \times \text{Rate} \]
\[ \text{Cost} \approx 1411.2 \times 0.50 = \text{Rs. } 705.60 \]
In simple words: Find the area of one triangular tile first, then multiply by 16 to get the total area of all tiles (approx. 1411.2 sq. cm). Since polishing costs 50 paise per sq. cm, the total cost is Rs. 705.60.

Exam Tip: Pay attention to unit conversions. Since the rate is given in paise (50 paise), convert it to Rupees (Rs. 0.50) before completing your final cost calculation.